Prove that for $A:= \{\operatorname{frac}(n\sqrt2)\mid n \in \mathbb{Z}\} $ we have $0.99 \leq \sup(A) \leq 1$

Question

Let $\operatorname{frac}(x)$ be the decimal part of a non‐negative real number. And $A:= \{\operatorname{frac}(n\sqrt2)\mid n \in \mathbb{Z}\}$. How can I show that $A$ is a bounded set and $0.99 \leq \sup(A) \leq 1$. Until now we have only the axioms of the real numbers and the supremum-infimum concept.

Answer 1

As $\operatorname{frac}(x)<1$ for all $x$, clearly $\sup A\le 1$.

Divide $[0,1]$ into $100$ intervals of length $0.01$ each. Then one of these subintervals must contain at least two of the numbers $\operatorname{frac}(n\sqrt 2)$, $n=1,2,3,\ldots, 101$. Note that $n\ne m$ implies $\operatorname{frac}(n\sqrt 2)\ne\operatorname{frac}(m\sqrt 2)$ because otherwise $n\sqrt 2-m\sqrt 2$ would be an integer and $\sqrt2$ would be rational! Therefore, we find $N,M$ with $$0<\underbrace{\operatorname{frac}(N\sqrt 2)-\operatorname{frac}(M\sqrt 2)}_{=:\delta}<0.01.$$

We find infinitely many (hence arbitrarily large) integers $n$ with $\operatorname{frac}(n\sqrt 2)>\sup(A)-\delta$. In particular, there eexist such $n$ with $n>M$. Then $$\operatorname{frac}((N+n-M)\sqrt 2) =\begin{cases}\operatorname{frac}(n\sqrt 2) + \delta& \operatorname{or}\\ \operatorname{frac}(n\sqrt 2)+\delta-1\end{cases}$$ (because $\operatorname{frac}(x+y)=\operatorname{frac}(x)+\operatorname{frac}(y)$ or $=\operatorname{frac}(x)+\operatorname{frac}(y)-1$). The top variant would make $\operatorname{frac}((N+n-M)\sqrt 2)>\sup(A)$, which is absurd. We conclude that the lower variant holds, which makes $$\sup(A)\ge \operatorname{frac}(n\sqrt 2) =\operatorname{frac}((N+n-M)\sqrt 2)+1-\delta>1-0.01=0.99$$

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Or in short: $$\operatorname{frac}(70\sqrt 2) =\operatorname{frac}(98.994949366\ldots)=0.994949366\ldots >0.99$$

Answer 2

It is true for any irrational number $\alpha$ that for $A:=\{\{n\alpha\}\mid n\in \mathbb Z\}$ (where $\{x\}$ denotes fractional part of $x$) we have $\sup(A)=1$.

Proof: Obviously $\sup(A)\leq 1$.

Observe first that if $n\neq m$ then $\{n\alpha\}\neq \{m\alpha\}$, otherwise $(n-m)\alpha=n\alpha-m\alpha$ would be an integer contradicting the irrationality of $\alpha.$

Let $\varepsilon:=\inf\{|\{n\alpha\}-\{m\alpha\}|\mid n,m\in\mathbb{Z}\}$. Observe that every one of the buckets $[i\frac\varepsilon 2,(i+1)\frac\varepsilon 2)$ containes at most a single $\{n\alpha\}$ (for $n\in\mathbb Z$) by definition of $\epsilon$. Since no two $\{n\alpha\}$ can be equal (see above) we must thus have infinitely many buckets, which means $\varepsilon=0$.

Now for every $\delta>0$ we can take $n,m=n+i$ such that $\{i\alpha\}=\{n+i\alpha\}-{n\alpha}<\delta$. Then if we look at the numbers $$\{n\alpha\}, \{(n+i)\alpha\},\{n+2i\},\dots$$ they get bigger and bigger in increments $\{i\alpha\}<\delta$. At some point we will therefore get within distance $\delta$ of $1$ which proves $\sup(A)\geq 1-\delta$.

Since $\delta>0$ was arbitrary we proved $\sup(A)=1$